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How do you differentiate #y=(x+1)/(x^3+x-2)#?

Answer 1

Samantha Adkison

Below

#y = (x+1)/(x^3+x-2)#

#(dy)/(dx) = ((x^3+x-2)(1)-(x+1)(3x^2+1))/(x^3+x-2)^2#

#(dy)/(dx) = (x^3+x-2-3x^3-x-3x^2-1)/(x^3+x-2)^2#

#(dy)/(dx) = (-2x^3-3x^2-3)/(x^3+x-2)^2#

The quotient rule is given by:

#y=u/v#

#(dy)/(dx) = (vu'-uv')/v^2#

In this case, your u is #x+1# and your v is #x^3+x-2#

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Answer 2

Luke Alvarez

To differentiate the function ( y = \frac{x + 1}{x^3 + x - 2} ), you can use the quotient rule of differentiation.

The quotient rule states that if you have a function in the form ( \frac{u}{v} ), then its derivative is given by:

[ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} ]

Where ( u' ) represents the derivative of ( u ) with respect to ( x ), and ( v' ) represents the derivative of ( v ) with respect to ( x ).

So, applying the quotient rule to ( y = \frac{x + 1}{x^3 + x - 2} ), we get:

[ y' = \frac{(1)(x^3 + x - 2) - (x + 1)(3x^2 + 1)}{(x^3 + x - 2)^2} ]

Now, simplify the expression in the numerator and denominator as needed.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.